The h Bar

0celo7

12:01 AM

@ACuriousMind So that rain turned into 5 inches of nasty, wet snow.

ACuriousMind

@0celo7 Nice...

Stan Shunpike

Does parallel transport mean relative to the manifold or the ambient space? But if the former, what does that mean? I always envision parallel as being defined relative to a plane of some sort. So I'm having trouble visualizing more complex structures where you might parallel transport

ACuriousMind

@StanShunpike I'd say it's parallel along the curve you are transporting, i.e. to someone travelling along the curve, the parallel transported vector doesn't turn, shrink or stretch

Stan Shunpike

No turn (direction), shrink or stretch (magnitude) so does that mean we define parallel using constant direction and magnitude.

ACuriousMind

Ehhhhh

0celo7

12:13 AM

@StanShunpike It's a little more complicated because the manifold has humps and bumps.

ACuriousMind

I'd say that's the intuition behind the definition, but not the definition

Because, for weird connections, the formal "parallel transport" doesn't look like a parallel transport at all

Stan Shunpike

Yeah, the humps and bumps are where I'm confused

0celo7

@ACuriousMind What does Mr. Germany like to drink?

ACuriousMind

@StanShunpike Have you looked at the equivalence of parallel transport and connection? Essentially, you are defining what it means to be "parallel" when you give the notion of covariant derivative - or you are defining what it means to covariantly derive when you give the notion of parallel transport

0celo7

@StanShunpike The definition of parallel transport is the geodesic equation.

@ACuriousMind I explained that in one of his Math.SE questions.

@StanShunpike I don't know the answer to this question.

ACuriousMind

12:19 AM

@0celo7 Well, unsuprisingly, all sorts of beer, but I also like gin and whiskey.

0celo7

@ACuriousMind Martinis are pretty awesome.

I put a little more Vermouth in mine than is considered normal, however.

(1:6 is standard, I like 1:4.)

ACuriousMind

Uggggh. I'll take a gin tonic over a martini any time :P

0celo7

Currently enjoying a nice single malt that my dad found.

No clue if it's any good :D

Seems alright to me.

Stan Shunpike

What confuses me is aren't there many kinds of connections? Whereas constant direction and magnitude are intuitive, I don't understand what the properties of a connection (for example as defined in Wald or do Carmo) has to do with commecting tangent spaces

ACuriousMind

@StanShunpike Well, because a connection defines parallel transport, it gives you a way to compare/connect vectors at different points.

0celo7

12:29 AM

@ACuriousMind We call it a connection on the tangent bundle because it connects the tangent spaces, right?

ACuriousMind

Another point of view is that it allows a unique lift of curves on the manifold into the bundle with connection.

@0celo7 What meaning of connect do you have here? (Also, the concept of connection is more general than tangent spaces)

(I don't really know what the original reason for calling it a connection is)

0celo7

@ACuriousMind I'm assuming you can define a connection over a general fiber bundle?

Danu

@vzn Oh sorry, I was just asking Daniel, and referencing your post

ACuriousMind

@0celo7 Yes

On general bundle, a connection is something like "choosing a horizontal/parallel direction at every point"

Sofia

@ACuriousMind isn't it bed-time for you?

0celo7

12:34 AM

@ACuriousMind Does "affine connection" refer to the covariant derivative connection?

Stan Shunpike

And I just learned from Wald that the uniqueness property of the Levi-Civita is crucial because it allows us to pick out a specific directional derivative operator. Or something like that. Why wouldn't there automatically be a unique on on a smooth manifold?

0celo7

@StanShunpike Unique what?

Stan Shunpike

One*

ACuriousMind

@0celo7 Well, you call any derivative induced by a connection "covariant".

Affine refers to the fact that the bundle fibers are affine spaces

0celo7

@ACuriousMind What is the "affine connection" then?

I meant the standard covariant derivative.

Stan Shunpike

12:37 AM

@0celo7 directional derivative operator.

I think that's the term he uses. I can go look

Danu

@ACuriousMind Yeah, looking at some books it appears that that's true. I think it makes no sense though

I think that vectors in Euclidean 3-space with the cross product should be considered a ring

ACuriousMind

@0celo7 There is no "the affine connection", but an affine connection is just a connection on an affine bundle (which the tangent bundle, as a vector bundle, is)

0celo7

@ACuriousMind Too many bundles! What is an affine bundle?

ACuriousMind

@0celo7 Heh. A bundle whose fiber is an affine space

0celo7

@ACuriousMind :/

Stan Shunpike

12:39 AM

Yay. I finally understand that.

ACuriousMind

An affine space is essentially just a vector space where you've forgotten that you've got an origin

0celo7

I know what an affine space is.

ACuriousMind

@Danu Why?

0celo7

What's the ramification of not having an origin?

Danu

@ACuriousMind Because it's one of the best-known examples of 'simple' multiplicaton of well-known objects

Based on my extremely small amount of experience with rings, it appears to me that the essence is 1) Abelian group w.r.t. addition 2) Multiplication plays nice with addition

so no constraints on multiplication per se

ACuriousMind

12:41 AM

@Danu But because it lacks associativity, it is not a good example - you can't easily speak about its ideals, its module theory is ugly, etc.

Danu

@ACuriousMind I guess I'm influenced by Vinberg :P

Stan Shunpike

If the affine connection is defined on an affine bundle, how does this allow us to measure curvature on a manifold?

ACuriousMind

And lacking unity, the allowed morphisms are so unconstrained because you don't have to send 1 to 1!

Stan Shunpike

Like why is it on the bundle as opposed to the manifold itself

ACuriousMind

It's really ugly from the viewpoint of most algebraic subdisciplines, I think

0celo7

12:43 AM

@ACuriousMind What are the consequences of not having an origin?

Danu

@0celo7 Greater symmetry group and stuff

ACuriousMind

@0celo7 Uhhh. Essentially, that you don't have an absolute notion of distance, only a relative one

(Though distance is handwavy here)

Sofia

@dmckee did you see this?

@dmckee shall I vote to close?

ACuriousMind

@StanShunpike Heh. Well, because every bundle has its own curvature, and what you usually call "curvature of the manifold" is really "curvature of the tangent bundle with the Levi-Civita connection".

And because the tangent bundle is somewhat natural to consider for a manifold, one does not really distinguish here

@Danu Well, but, for example, a very interesting construction is localizing a ring, i.e. partially making its elements invertible w.r.t. multiplication. This fails as a nice construction horribly if you don't have associativity, and you can't even say what invertible means lacking a unit

Sofia

@dmckee it has smth. to do with the Green function, s.t. I am not sure.

Danu

12:48 AM

@ACuriousMind Well, you just have to say "associative ring with unity" then

ACuriousMind

@Danu Yes. But I've never seen any interesting results about non-associative ring without unit

So it is more economic to generally assume associativity and unity when someone says ring

That's not to say that there aren't results about the more general rings, but they don't show up that often

Danu

I'll get back to you when I read Vinberg's chapter on rings ;)

oh jk, he doesn't have one

ACuriousMind

Huh? :D

Danu

It's not a very big book (sub 500 pages)

ACuriousMind

B-b-b-but...what do you do in algebra if not rings?

500 pages of group theory?

Or Galois stuff?

Danu

12:54 AM

Affine & Projective spaces, Tensor algebra, commutative algebra (I guess this is largely about rings?), groups, representation theory, lie groups

Thats chapters 7-12

ACuriousMind

Affine and projective spaces without rings? Oh, well...

But yeah, commutative algebra should be module theory over commutative rings, essentially

How is that just a chapter?

Sean

0

Q: Does Big Bang Cosmology imply/require infinite space?

The reason I am asking this question is because if all points in space observe recession of galaxies the same as we do from Earth, the universe would have to be infinite (or a closed sphere in 4D or something. I know infinite space isn't a formal position of Big Bang cosmology, but is a non infi...

Danu

It's not a very advanced book, so that's probably why?

Sean

The Big Bang does not_require_ that space is infinite, it just so happens that that's the case. Right?

ACuriousMind

@Danu I see. I can then only predict that you'll have to learn everything all over again if you ever need to go into more depth in any of the subjects in there.

Danu

12:57 AM

@ACuriousMind I think it comes quite highly recommended by math.se users so I don't think it is bad.

Stan Shunpike

How can curvature of a bundle inform me about curvature of a manifold? If I am measuring properties about the bundle, is there a one-to-one correspondence between points in the bundle and points on the manifold such that I get info about the manifold from the bundle?

Danu

Also, the first chapter already introduces the most basic notions so the dfinition of rings fields etc is in there.

ACuriousMind

@StanShunpike The Levi-Civita connection is uniquely determined by the metric, so it carries unique information about the manifold.

@Danu Well, we'll see what you say after you've read it. I can't judge its quality from chapter titles, I guess

Danu

8

A: Good books for self-studying algebra?

The absolute best book for self-study in algebra to me is E.B. Vinberg's A Course In Algebra. A Course In Algebra by E.B.Vinberg This book very rapidly became my favorite reference for algebra. Translated from the Russian by Alexander Retakh, this book by one of the world’s preeminent algebracist...

this is where I got it from

Stan Shunpike

@ACuriousMind the connection is defined on the bundle. And the tangent bundle is just the disjoint union of tangent spaces. So what does that have to do with the original manifold? Like if the tangent bundle has the structure of a connection that doesn't mean the manifold does, right?

ACuriousMind

1:09 AM

@StanShunpike The manifold cannot have the structure of a connection, because connections, by definition, live on bundles. "Disjoint union" is a characterization of the tangent bundle I've grown to dislike, because it does not carry the disjoint union topology.

I'd rather say that the tangent bundle arises naturally as the vector bundle on which the Jacobians of the coordinate transformations between the local charts of the manifold act, but for this to make sense, you'd have to internalize that bundles are defined by local patches and transitions between them.

Stan Shunpike

Hmm...I see your point. I thought that definition seemed oversimplisitc somehow. Why do they have to live on bundles? Wikipedia does say that is part of the definition but doesn't say why. wikipedia mentions a space of infinitely differentiable vector fields and the connection is a map on this. Can that space not be defined on the manifold itself without the bundle?

I thought smooth manifolds were by definition infinitely differentiable.

ACuriousMind

@StanShunpike If you define "that space" on the manifold, the bundle is what you get

That's what the tangent bundle is - it's where the tangent vectors live

Stan Shunpike

Ohhhh

ACuriousMind

And because a connection is a notion of shoving around tangent vectors, it also lives on the bundle

Stan Shunpike

1:47 AM

@ACuriousMind One of my earlier questions was, what are we looking for that the manifold doesn't already have?

Like, the manifold doesn't allow us to by itself compare tangent spaces at points

But how do we know what properties we need in order to compare them?

@Danu its a shame topos is so hard. I really was hoping he has some good ideas.

music theory sux. I haven't found any practical ones so far. Most of it I make up.

NeuroFuzzy

@Danu That's a really good infomercial for that book

Stan Shunpike

2:56 AM

On page 31 of Wald, he discusses defining a commutator of two vector fields "in terms of any derivative operator". What does he mean? Does he mean write the vector fields in terms of the derivative operator or the commutator?

Kyle Kanos

4:13 AM

Am I wrong in saying that this really isn't much of an answer?

Nick

7:44 AM

I shudder at the offtopicness of math in this room. Smoking should be done in the smoking room. It's annoying if it isn't.

:D With that said, it's great that everyone here is so well-rounded.

@KyleKanos The answer is in the right direction. There's a huge lack of descriptiveness.

DanielSank

@Danu Yes!

My PhD work was an important part of getting that experiment to work :)

Nick

I have a tiny little question which isn't even worthy of main. Maybe I could ask it out here.

From my understanding of Nulear Fission

^ A heavy nucleus is excited and split into two smaller nuclei.

A while ago, one of my teachers commented that the combination of $^3_2\text{He}$ in the sun to form $^4_2\text{He}$ is fission.

Then, someone in the class said it was fusion. And an unintelligent debate ensued where one party screamed fission, the other fusion.

I know this seems silly but I too am wondering what the following is:

$$^3_2\text{He}\ +\ ^3_2\text{He}\ \rightarrow\ ^4_2\text{He}\ +\ ^1_1\text{H}\ +\ ^1_1\text{H}\ +\ 12.8\text{MeV}$$

All I'm familiar with are $\alpha, \beta, \gamma$ decays and the above is no direct disintegration.

I too feel that this is thermonuclear fusion. But maybe my lack of knowledge on this subject leaves me out of being any judge of it. So, could someone please inform me on what the mechanism behind the above is?

... oh wait! Ive found it

It's in the wiki article for fusion.

Ohk, it's $\beta^+$ decay followed by $\gamma$ decay and then fusion.

So, I guess it's still fusion.

What was my teacher smoking to have said fission!?

The problem here was, I think, a lack of insight into how that final step took place. He must have thought that the heliums combined together and split up again as opposed to a head on collision during which two protons are thrown away.

Danu

8:48 AM

@DanielSank So... do you have a PhD from 'Google Uni.'? ;)

Nick

8:59 AM

@Danu No, but I think I deserve one. I grew up on it.

It's still hard to find all the answers though.

Danu

LOL

Danu

9:23 AM

@NeuroFuzzy That's what I thought, lol

Stan Shunpike

I don't understand what rings are useful for

Whereas fields seem naturally useful, rings seem like a clunkier version without the nice features of being abelian. Are rings used a lot in physics?

Danu

@StanShunpike You can have commutative rings that are not fields (I assume that's what you mean by abelian, which is normally just used for groups)

ACuriousMind

10:29 AM

@StanShunpike Hm...well, we do define the parallel transport to transport the vectors around, and then we can compare them, right?

@StanShunpike "Useful"...what is that? ;)

@KyleKanos No

ACuriousMind

11:51 AM

Is this a joke? OP says "I'll make my question more specific" and then does a edit that does...absolutely nothing

0celo7

12:15 PM

@ACuriousMind Ray Kay is a resident troll, you know.

@ACuriousMind Blumenhagen, Basic Concepts in String Theory (2013) says that if the Hamiltonian vanishes, then the dynamics of the system are determined by the primary constraints. I don't remember if that's what you said.

ACuriousMind

@0celo7 In a way, it is, I said that the Hamiltonian vanishes on the constraint surface

So, indeed, the e.o.m are completely determined by the constraints

0celo7

@ACuriousMind Fancier language.

tpg2114

!@$*(&^KJHSADJKHD ARRRRRGGGGGHHHHHHHHH. I just got f'ing scooped again. Makes me so mad!

sciencedirect.com/science/article/pii/S0045793015000092

I have those damn simulations with the same damn new formulation sitting in the f'ing queues waiting to run

0celo7

@tpg2114 TFW high school peasant and no access to journals...

@ACuriousMind Eigenstate measurements aren't probabilistic in nature, are they?

tpg2114

@0celo7 Just had to look up TFW... kids these days, can't understand anything they say

ACuriousMind

12:22 PM

@0celo7 Blumenhagen is even a bit more precise because he notices that there cannot be secondary contraints

@0celo7 Well, I'd say they are, just with probability 1 ;)

But you can quibble over language there

0celo7

"does the spin of the particle depend upon probability"

???

ACuriousMind

@tpg2114 Again? How often have you suffered this now?

tpg2114

If you want open access reading material, Phil. Trans. A has an open access edition of commentary on some of the most important papers published in it: rsta.royalsocietypublishing.org/content/373/2039?etoc

0celo7

Quantum Entanglement & Spooky Action at a Distance

►

This guy...

tpg2114

@ACuriousMind Maybe 4-5 times in the past 6 years.

This one at least I'm still waiting for data. There have been other times that we're finishing our paper to submit and a virtually identical paper shows up as an early access article

0celo7

12:28 PM

@ACuriousMind His question is an exercise in Shankar.

Which my Psych teacher gave back, BTW.

Guess he was overconfident in his ability to do real math.

ACuriousMind

@0celo7 dafuq did I just watch

The spinning guys are...distracting :D

0celo7

@ACuriousMind It's not wrong, I don't think.

ACuriousMind

No, it's actually not that bad

The video looks as if there were drugs involved, though

0celo7

Ha

Very noice Bell's theorem intro.

ACuriousMind

I think it is really one of the best explanations I've heard

And he actually also explains no-communcation

I think I like this guy

bolbteppa

12:46 PM

Guys this whole Hamiltonian vanishing thing is driving me insane

Why did I peek under the covers...

0celo7

@bolbteppa Did you read my chat post a few minutes ago?

bolbteppa

about Blumenhagen?

0celo7

Yes.

BTW this book is pretty good. It's (exactly) like BBS but with more hand holding.

bolbteppa

I got that part, but it's much deeper than all this tbh

ACuriousMind

@bolbteppa Yes, it is. I advise to read about constrained Hamiltonian systems

bolbteppa

12:52 PM

I know it's dealing with all the Dirac bracket stuff, it's in Weinberg and Dirac's Yeshiva lectures and a perimeter video lecture, but why do we have constrained Hamiltonian systems in the first place?

ACuriousMind

Two main possible reasons:

1. You started with a Lagrangian whose Legendre transform is degenerate

2. You have a generally covariant system, i.e. invariance under time-reparametrisation

Here, 2. gives you a constrained system if you describe the system with time and space on equal footing, and all variables depending on an abstract "evolution parameter" that need not be time

While 1. essentially means you had gauge symmetry to begin with

0celo7

@ACuriousMind Is string theory both?

ACuriousMind

@0celo7 Yes, string theory is both, I believe

1. and 2. are not actually different from the Hamiltonian point of view - if you just always go to a time-space symmetric description, you cannot tell the "internal" gauge constraints from the "spacetime" gauge constraints

0celo7

I hate constraints. Dirac brackets are pretty terrible.

ACuriousMind

I have learned to appreciate their beauty through the book I'm reading ;)

bolbteppa

12:59 PM

@ACuriousMind if you check out Rund's "The Hamilton-Jacobi Theory in the Calculus of Variations" he shows how the Hamiltonian function will not exist for any 'homogeneous' action functional, where homogeneity means parameter invariant, and he shows how to get a Hamiltonian as a kind of inverse metric

I think this book has a proper explanation of the Dirac bracket stuff, without using those words, it's more geometric

ACuriousMind

@bolbteppa: I believe he does no do that in fully generality, because there are two caveats to the statement "The Hamiltonian vanishes for generally covariant systems":

1. The Hamiltonian only vanishes weakly, i.e. it may be non-zero off-shell

0celo7

@ACuriousMind You quantum theorists and your off-shellness.

ACuriousMind

2. If you choose phase space coordinates that do not transform as scalars under time reparametrisation, then you also can get non-zero Hamiltonian

Nevertheless, there will always be gauge freedom that allows you to reach vanishing Hamiltonian

@0celo7 off- and on-shell are perfectly classical concepts :P

They just aren't that useful/relevant classically as they are quantumly

0celo7

@ACuriousMind No engineer has ever worried about being on-shell or off-shell.

ACuriousMind

@0celo7 I doubt engineers do Hamiltonian or Lagrangian mechanics beyond solving the e.o.m.

0celo7

1:06 PM

@ACuriousMind I never said they did any more.

bolbteppa

I'm not sure about that, he literally just uses Euler's theorem on homogeneous functions to prove that the Hamiltonian will always vanish, because differentiating $\mathcal{L}(x^i,a \dot{x}^i) = a \mathcal{L}(x^i,\dot{x}^i)$ w.r.t. a gives $\frac{\partial \mathcal{L}}{\partial \dot{x}^i} \dot{x}^i = \mathcal{L}$ which means $H = \frac{\partial \mathcal{L}}{\partial \dot{x}^i} \dot{x}^i - \mathcal{L} = 0$ identically regardless of whether the EOM are satisfied or not

ACuriousMind

@0celo7 So, your statement is not an argument against my statement that "shellness" is a classical concepts. Engineers != Classical physicists

0celo7

@ACuriousMind I wasn't arguing against classical physics, just that to an engineer, the distinction between on-shell and off-shell is nonsense.

In my mind, engineers are intellectually impoverished "good 'ol boy" classical physicists.

(Except nuclear engineers. They have to take QM classes.)

ACuriousMind

@bolbteppa Given time $t$ and evolution parameter $\tau$, define $T = t(\tau) - \tau$ and write down the full Hamiltonian action. It is generally covariant, but does not imply vanishing of the Hamiltonian. The issue is that the Hamiltonian (at least, the thing that is in the action) is, for constrained systems, not just the Legendre transform - you have to add Lagrange multipliers for the constraints for the actual thing that appears in the action

There are a lot of subtleties in these constrained systems

0celo7

@ACuriousMind Oh Lagrange multipliers...

ACuriousMind

1:14 PM

BTW, I'm not pulling these statements out of my behind, this is chapter 4 in Quantization of gauge systems

@0celo7 What about them? Don't you like them?

0celo7

@ACuriousMind They seem like magic, but I'm sure they're rigorous. Like $i\epsilon$.

bolbteppa

So lets get this straight, if you have an arbitrary action functional, you can define the Hamiltonian function as the Legendre transform so long as some convexity second derivative condition is satisfied. If that fails, then for arbitrary actions this theory is a nightmare plagued with research questions (as my professor said). One special case in which the convexity argument fails is the case of a homogeneous action (e.g. the SR action). There are two reasons for this:

0celo7

@ACuriousMind I'm finally getting to the point in string theory where I can quantize the bosonic string!!

Yay I know chap 1!

bolbteppa

1) Homogeneous Lagrangians are parameter independent, which means parametrizing it by some other variable shouldn't affect the form of your action
2) The Legendre transform is actually just a change of variables, a change of parametrization from point-point coordinates to line-point coordinates, generating the surface by points and lines instead of just points (obviously the (line,point) coordinates are points in some new space)

Wow

!!!

That's the physical intuition right there!

0celo7

@ACuriousMind When we do PBs wrt. fields and momentum densities, we have to integrate over speacetime, right? Or, in string theory, do we integrate along the string?

bolbteppa

1:27 PM

Because conservation of energy arises from invariance of the action under changes in the time coordinate, just because the Legendre transform gives zero doesn't mean energy isn't conserved

0celo7

@ACuriousMind i.e., how does $$\{X^\mu(\sigma,\tau),\Pi^\nu(\sigma',\tau)\}_\text{P.B.}=\eta^{\mu\nu}\delta( \sigma-\sigma')$$ work?

ACuriousMind

@0celo7 I don't think we integrate, that's why the Dirac deltas are there

But your coordinates are $\sigma,\tau$. I don't think you ever integrate over target space in any case

0celo7

@ACuriousMind Don't we get a factor $\delta( \tau-\tau)=\delta(0)$ because they are equal time?

ACuriousMind

Ah...well.

Probably

And we probably don't care :D

But I'm not sure

0celo7

::gets out Weinberg::

I must have been on drugs when reading this.

There were definitely integrals.

AHA

p. 347 the first unnumbered equation

We integrate if we are taking the P.B. of a functional.

bolbteppa

1:35 PM

Yeah!

0celo7

@ACuriousMind $$\frac{1}{2\pi i}\oint dz\, z^{n-1}=\delta_{n,0}$$ right?

ACuriousMind

@0celo7 Yes

0celo7

@ACuriousMind I can also derive the Virasoro algebra then.

ACuriousMind

Yes :D

bolbteppa

However, because the Lagrangian is parameter independent, it is constrained in the choices of coordinates $(x^i,\dot{x}^i)$ that one is allowed to plug into it, constrained so that we always preserve homogeneity, hence a homogeneous functional is equivalently described as a constrained variational problem, but written out this way we return everything to normal and can use Legendre transforms :)

0celo7

1:40 PM

@ACuriousMind My mistake was buying my first string theory book by only looking at Amazon reviews. BBS is not introductory.

It's like trying to learn GR from Wald.

bolbteppa

In Rund's book there is an equivalent way to determine the Hamiltonian directly from the constrained functional and not when it's written out as a constrained variational problem, motivated by Finsler geometry"

My guess is that Dirac brackets are the lie algebra method of performing the Legendre transform in the tangent space of the constrained system, and the stuff above is just the Lie group method of staying in the constrained tangent space, whether directly (Rund) or using Lagrange multipliers

@ACuriousMind I don't understand how you stopped the Hamiltonian vanishing by just choosing a special time variable

ACuriousMind

1:57 PM

@bolbteppa We start from the generally covariant Hamiltonian action $$\int p_0 \dot{q}^0 + p_i\dot{q}^i - u^0(p_0 + H_0) \mathrm{d}\tau$$, where $u^0$ is the Lagrange multiplier enforcing the vanishing of the total Hamiltonian, which is just all terms which are not $p\dot{q}$. If you now choose the time coordinate as I did, you generate an additional summand $H_0$, so the total Hamiltonian becomes non-zero

It is interesting to note that this is not special to time reparametrization invariance

bolbteppa

That is equation 4.2 in Henneaux (in a special case), but you can only write that after you set up equation 4.1 as a constrained variational problem and reduce it to 4.2, more generally you get 4.6a after turning 4.1 into a constrained variational problem, and only once you've done this can you use 4.9, but again that's perfectly explained by what I was saying! He defined a change of variables that was not a reparametrization but something that transforms as a connection (4.11).

ACuriousMind

I did not say it was a reparametrization, did I?

It's an allowed change of variables, though

The remark after 4.11 is interesting because it shows vanishing Hamiltonians always occur as soon as you have suffiently nice gauge freedom

lol...I almost retagged something with hamiltonian-formalism instead of homework-and-exercises because of this chat

0celo7

@ACuriousMind Who is Qmechanic anyway? What's his specialty?

ACuriousMind

@0celo7 Nobody knows :)

0celo7

@ACuriousMind The Qmechanic uncertainty principle?

@ACuriousMind Is $$\frac{1}{\ell}\sum_n \exp(2\pi i n\sigma/\ell)=\delta(\sigma)$$ just the Fourier series for the delta function?

ACuriousMind

2:09 PM

@0celo7 Looks like it, yes

@0celo7 Heh :D

0celo7

@ACuriousMind Fourier series are not that common in physics.

Laplace transforms are very rare.

ACuriousMind

Well, Fourier series become common when you are in a finite, periodic setting, like the string

I have no idea what a Laplace transform is off the top of my head, though

0celo7

@ACuriousMind $$\mathcal{L}(f)(s)=\int_0^\infty e^{-st}f(t)\,\mathrm{d}t,\quad s\in\mathbb{C}$$

Notation on the left might be funky.

bolbteppa

So to sum up again, given a homogeneous action we know that the naive Legendre-transform Hamiltonian will be zero, because a Legendre transform is just a reparametrization. However because one must preserve homogeneity we know the problem is equivalent to a constrained variational problem also. Once we set up the constrained variational problem, we can keep parameter invariance and end up with a Hamiltonian, but the change of variable in the parameter no longer change as a scalar

0celo7

@ACuriousMind Engineers use it for everything apparently.

ACuriousMind

2:14 PM

@bolbteppa Yeah, I think that's correct

@0celo7 ::shrug::

0celo7

@ACuriousMind We should all adopt Qmechanic's $\approx$ notation. String theory is horrible without it. Also, there should be an equals sign for "equal under an integral sign". Stuff like $x^2\delta'(x)=0$ is atrocious.

ACuriousMind

@0celo7 I fully agree.

0celo7

@ACuriousMind Or "equal when in a correlation function". CFT and OPEs would be a lot easier to understand, if only a bit more cluttered.

Catchy.

The Fuze - Hybrid #TiH

►

@ACuriousMind I'm stuck on deriving the Witt algebra. I know that I have to calculate the P.B. of the energy-momentum tensor with itself and then Fourier expand. I just don't know how to do the P.B.

Nick

2:39 PM

Hey, guys. In the thin lens version of the lensmaker's formula

$user image$

What does $n$ stand for?

ACuriousMind

@0celo7 Uffff, I'm not sure I've ever done that explicitly myself. I think I didn't

bolbteppa

@Nick en.wikipedia.org/wiki/Lens_%28optics%29#Lensmaker.27s_equation

@0celo7 it's in Blumenhagen's Conformal Field Theory book page 13

Nick

@bolbteppa Yes, it says is the refractive index of the lens material. But I think it's wrong in a medium such as water.

0celo7

@bolbteppa I know how to do it that way.

ACuriousMind

@Nick You probably have to replace the $1$ in $n-1$ there by the refractive index of the surrounding medium

bolbteppa

2:42 PM

@0celo7 books.google.ie/…

0celo7

I don't want to do it that way though.

bolbteppa

What are you trying to do then? That's the only way I've ever seen it done, where are you reading about it?

0celo7

@bolbteppa Blumenhagen's string theory book ;)

bolbteppa

haha

Nick

@ACuriousMind OR I can leave the equation as it is and say that $n$ is the relative refractive index of the material with which the lens is made with respect to the medium in which it stands. Should I add that in wiki?

0celo7

2:43 PM

p. 26

ACuriousMind

@Nick Probably. I'm no optics guy, though

0celo7

I think I skipped a pedagogically important calculation earlier on.

ACuriousMind

(Not an optics girl, either)

0celo7

@ACuriousMind is an it.

Nick

@0celo7 Like from the adam's family?

..oh wait, that was the Thing.

ACuriousMind

2:45 PM

@0celo7 Not an optics it, either, then :D

@Nick Oh, the hand?

bolbteppa

I think you just plug equation 2.83 into 2.88

ACuriousMind

It's called das Eiskalte Händchen in the German translation. Is it really just "the Thing" in the original?

bolbteppa

and it's the same idea as in his CFT book more or less, but I have to sit down and find the direct link asap myself

Nick

@ACuriousMind Yes, the hand. I like the hand. I remember a scary movie where a guy's hands developed a conscience of their own and started killing people. He cut them off. Do you know the name of that movie? I think it could be from the stories from the crypt.

0celo7

@bolbteppa You plug 2.85 into 2.89 to get 2.88

I want to get the first of 2.89

No real clue how to do it.

Nick

2:47 PM

On a similar note, conscience is a strange word to spell. con-science ... fake science?

ACuriousMind

@Nick Don't know that, I only watch horror movies if they're hilariously bad :D

@Nick It's con- from the latin prefix cum- for "with, together with"

bolbteppa

I think you can do it both ways and my way is easier

Nick

@ACuriousMind It was hilariously bad but no scary movie (the five part trilogy)

0celo7

@bolbteppa I know your way is easier. Doesn't mean I shouldn't be able to do it this way too.

bolbteppa

I think I have to do a similar crazy calculation for my professor involving the Heisenberg algebra haha

0celo7

2:50 PM

I think I have to write the momentum in terms of light cone derivatives.

Then chain rule?

Nick

@ACuriousMind Conscience is "together with science"? Hey, that's my middle school textbook.

ACuriousMind

@Nick Well, scientia is knowledge

So, conscientia was that which came with knowledge

Nick

@ACuriousMind "together with knowledge" ... mhh, it screams "rationalism"

ACuriousMind

Well, your conscience is the thing that nags you because of the knowledge of your past deeds. I think it's not at all referring to scientific knowledge as we understand it

Nick

@ACuriousMind I think sociapathic tendancies follow the most knowledgable. It takes wisdom and empathy to have a conscience.

ACuriousMind

2:52 PM

"scientific knowledge" is knowledgy knowledge literally, by the way :D

Nick

@ACuriousMind I feel I can learn Latin from you.

ACuriousMind

Yes, you probably could

0celo7

@ACuriousMind Well that's the best kind

Nick

@ACuriousMind All I know is Lorem Ipsum.

ACuriousMind

@0celo7 True enough

@Nick It's a beautiful language. They don't make them like that anymore ;)

Nick

2:56 PM

@ACuriousMind As I know, true Latin speakers are dead. Maybe we'll all try to speak it again one day. Illegitimi non-carborundum.

0celo7

@ACuriousMind TFW missing a delta function and no clue how to take the momentum derivative of the EM tensor.

ACuriousMind

@Nick My Latin teacher could speak fluent Latin. I can't, I can only read it quite well.

@0celo7 I hate chasing deltas

0celo7

I hate P.B.s.

Nick

@ACuriousMind How do you know it's fluent? There's no living native to compare it with!

@0celo7 but... (continue your sentance with a but, it makes things positive)

0celo7

@ACuriousMind Frick. Missing two deltas.

ACuriousMind

2:58 PM

@Nick Well, fluent as in: Able to translate a moderately long German text on the fly into Latin and speak it aloud during that.

0celo7

How does this even happen??

Nick

@ACuriousMind O_O Wow

0celo7

@Nick There is nothing positive about Poison brackets. (Typo intentional. These things are evil.)

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